Answer:
The car travels 36.8 m before coming to stop after the light changes
Explanation:
The car moves at a constant speed of 59.0 km/h for 0.750 s before the driver hits the brake.
The equation for the position of an object moving at constant speed is:
x = x0 + v t
where:
x = position at time t
x0 = initial position
v = speed
t= time
Let´s consider the initial position as the position at which the driver sees the traffic light and decides to brake. That will make x0 = 0. Then, the position after 0.750 s will be:
x = 59.0 km/h * 0.750 s (1 h /3600 s) = 0.0123 km (1000 m / 1 km) = 12.3 m
while braking, the car has a negative acceleration, then, the speed is not constant. The position of the car will be given by the following equation:
x = x0 + v0 t + 1/2 a t² ( where a = acceleration and v0 = initial speed)
and the speed can be expressed as follows:
v = v0 + a t
from this equation, we can calculate how much time it takes the car to stop (v = 0):
0 = v0 + a t
-v0 = a t
-v0 / a = t
v0 is the speed of the car as the driver hits the brake (59.0 km/h) and "a" is the acceleration (5.50 m/s²) that will be negative because the car is losing speed. Then:
-59.0 km/h (1000 m / 1 km) (1 h / 3600 s) / (-5.50 m/s²) = 2.98 s
Now, we can calculate the position at this time to know the minimum distance the car travels before coming to stop:
x = x0 + v0 t + 1/2 a t²
now x0 will be the distance traveled after the driver sees the light but before braking ( 12.3 m). v0 will be the speed before braking, 59.0 km / h or 16.4 m/s. Then:
x = 12.3 m + 16.4 m/s * 2.98 s +1/2 (-5.50 m/s²) * (2.98 s)²
x = 36.8 m
